Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s.

a. What is Sam's top speed?

Question

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s.

a. What is Sam's top speed?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to determine the top speed attained by Kevin. We will have Kevin having a mass of 65 kg riding a bicycle on a flat road. The bicycle has a force of 100 and 25 Newton pushing it forward and a coefficient of kinetic friction on the road of 0.8. The chain on a bicycle falls off after 25 seconds. And we're being asked to determine the top speed attained by Kevin. And the options given are a 3.98 m per second. B 1.59 m per second. C 10.3 m per second and D 11.4 m per second. So we will use the particle model and consider Kevin as a particle. The pictorial representation of the situation given in the problem statement will be de de depicted as follows. So I'm going to just start drawing the xny axis of our free body diagrams. So I'm going to draw two different free body diagrams. Uh The first one will be for um when Kevin is accelerating and the second one is for when Kevin is coasting just like. So, so when Kevin is accelerating, um the forces acting upon it will be first, there will be the gravitational acceleration or the FG gravitational force and the countering normal force. At the same time, there will be FK which is the frictional force countering the force pushing cabin forward, which is the um 125 Newton force. I'm just gonna to indicate it, indicate that with an F and when Kevin is coasting still, there will be normal and gravitational force. So N and FG, but there will only be friction force acting on DX axis. So it's going to only be FK, there will be no acceleration in the vertical direction for both when Kevin is accelerating and also when Kevin is coasting. So in both cases, we can use the new terms law of Sigma Fy equal to zero where Sigma Fy is just N minus MG or M minus FG. So in this case, we can then determine that that's an arrow N will be equals to M multiplied by G just like. So, so the normal force N can then be calculated to then B the mass of Kevin is 65 kg and the gravitational acceleration which is 9.81 m per second squared. So calculating that the normal force will then come out to a value of 637 Newton. Next, we can calculate the kinetic friction or FK which if you recall, it's just essentially UK multiplied by N the UK is given in the problem statement to be 0.18. And the normal force is what we just calculated, which is 637 new, calculating this, the friction force or kinetic friction of 100 and 4.7 Newton. Next, we want to apply Newton second law in the nap force in the X direction. So Sigma FX will not be equals to zero but equals to M multiplied by A X. And in this case, the only uh forces acting on the X direction when Kevin is accelerating is just the 135 Newton and the friction force. So let's do that. So Sigma FX is going to be 100 and 25 Newton minus the FK just like so which will be equals to M multiplied by A X and FK is just essentially 114.7 Newton. So this will be 1 25 Newton minus 1 14.7 Newton. And that will give us 10.34 Newton, which if we want to calculate the acceleration A X, we can then calculate that with 10.34 Newton divided by the mass of Kevin, which is 10.34 Newton divided by 65 kg, which will then come up to a value of acceleration to be 0.159 m per second squared. Next, we want to apply the kinematic equation of P one equals to P zero plus a zero T one where P one is the final velocity or the top speed, which is what we are interested to find. P zero is just the initial velocity which is going to be 0 m per second. And then a knot is essentially just A X which will be 0.159 m per second squared. Awesome. P one is just the time which is given to be 25 seconds in the problem statement. So let's substitute all of our information that we know. So P one is top speed. P no is zero uh plus a knot which is equals to 0.159 m per second squared. And T one is 25 seconds. Calculating this V one will come up to a value of 3.98 m per second, which will be the answer to this particular practice problem where the top speed of Kevin is the going to be equal to 3.98 m per second, which will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that'll be it for this one. And thank you.

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s. a. What is Sam's top speed?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Friction

Kinematic Equations